Can a Rubik's Cube be brute-forced?

I had a prof in college describe solving a rubik&#x27;s cube as group conjugation (e.g. f * g * f^-1).<p>For example, you find a way to swap two pieces on the top layer and mangle the bottom (f), turn the top (g), and then do the opposite (f^-1), swapping a different pair and un-mangling the bottom. Between complementary swaps, edge flips, and corner rotations, you can build an entire solution with this technique. (My current version of this does the edges first, ignoring any damage to corners and then does corners.)<p>Somewhat related - many years ago there was a tutorial of the Gap computer algebra system that analyzed the rubik&#x27;s cube group. I can&#x27;t find the original, but there is a translation to Julia here: <a href="https:&#x2F;&#x2F;oscar-system.github.io&#x2F;GAP.jl&#x2F;stable&#x2F;examples&#x2F;" rel="nofollow">https:&#x2F;&#x2F;oscar-system.github.io&#x2F;GAP.jl&#x2F;stable&#x2F;examples&#x2F;</a>

Of course the true brute-forced algorithm is to pry the cube apart and then assemble it again in the solved state.

Related:<p><i>Can a Rubik&#x27;s Cube be brute-forced?</i> - <a href="https:&#x2F;&#x2F;news.ycombinator.com&#x2F;item?id=36645846">https:&#x2F;&#x2F;news.ycombinator.com&#x2F;item?id=36645846</a> - July 2023 (108 comments)<p>(Reposts are fine after a year or so; links to past threads are just to satisfy extra-curious readers)

&gt; The essence of the result is this. Reminiscent of a “meet in the middle” algorithm<p>Hm, so something akin to a bidirectional path-finding problem, where one can still call it &quot;brute force&quot; because both known positions (start and goal) are each doing a breadth-first search, as opposed to something fancier than picks a direction.<p><a href="https:&#x2F;&#x2F;en.wikipedia.org&#x2F;wiki&#x2F;Bidirectional_search" rel="nofollow">https:&#x2F;&#x2F;en.wikipedia.org&#x2F;wiki&#x2F;Bidirectional_search</a>

Fun to see someone else look at the cube as permutations. <a href="https:&#x2F;&#x2F;taeric.github.io&#x2F;cube-permutations-1.html" rel="nofollow">https:&#x2F;&#x2F;taeric.github.io&#x2F;cube-permutations-1.html</a> is one of the more fun visualizations I&#x27;ve made that was looking at this. Reminds me I need to finish it. I can&#x27;t remember why I haven&#x27;t gotten further.

This reminds of the time I decided to teach one of my kids about computer programming, and suggested we write a program to solve a rubik&#x27;s cube as an example (rubik&#x27;s cubes were popular at her school at the time). Only after we had written a really simple depth-first search did I realize that it would take several lifetimes to run. Great lesson, but not the one I meant to impart!

Compare another &quot;brute force&quot; method: if we imagine the state space of all Rubik&#x27;s Cube configurations as nodes and quarter-turn transitions between them as edges, it turns out there exists [0] a Hamiltonian circuit for this massive graph. By symmetry, we can traverse this circuit with the same moves starting from any configuration.<p>Amazingly this means we can solve a Rubik&#x27;s Cube without ever knowing what the original configuration is, as long as we can ask at any time if the cube is solved or not.<p>[0] <a href="https:&#x2F;&#x2F;bruce.cubing.net&#x2F;ham333&#x2F;rubikhamiltonexplanation.html" rel="nofollow">https:&#x2F;&#x2F;bruce.cubing.net&#x2F;ham333&#x2F;rubikhamiltonexplanation.htm...</a>

I feel like a simple brute force method is using a &quot;labirinth&quot; exploration. Make a list of all possible moves in a single state, do one of these moves and check: if you already saw the result, go back and try a new branch until the cube is solved.

I’d say; Yes using only about 7 steps where each repeats a permutation, but it requires reorientation of the cube ofcourse.<p>But reassembling is the true brute.

Sure it can. At the peak of Cube popularity, I saw &quot;game set&quot; in store. Included Rubik&#x27;s Cube and hammer.

This sounds like bidirectional search + rainbow tables

After 6831 (give or take) hours, I&#x27;m pretty sure the answer is &quot;no&quot;